Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ 1-x+\frac{x^{2}}{2}-\frac{x^{3}}{3}+\frac{x^{4}}{4}-\cdots $$

Step-by-Step Solution

Answer

The convergence set is \([-1, 1)\).

Step 1: Identify the General Term

The given series is \( 1-x+\frac{x^{2}}{2}-\frac{x^{3}}{3}+\frac{x^{4}}{4}-\cdots \). We can recognize this as an alternating power series. The general term is \((-1)^n\frac{x^n}{n}\) where \( n \) starts from 0.

Step 2: Apply the Absolute Ratio Test

For a power series \( \sum a_n x^n \), the radius of convergence \( R \) can be found using the Absolute Ratio Test: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \)Substitute the general term: \[ a_n = \frac{(-1)^{n}x^n}{n} \] and \[ a_{n+1} = \frac{(-1)^{n+1}x^{n+1}}{n+1}. \]So, the expression becomes:\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(-1)^{n+1}x^{n+1}}{n+1}}{\frac{(-1)^n x^n}{n}} \right| = \left| \frac{x}{1 + \frac{1}{n}} \right|. \]

Step 3: Find the Limit for Convergence

Simplify the expression from Step 2: \[ \lim_{n \to \infty} \left| \frac{x}{1 + \frac{1}{n}} \right| = \lim_{n \to \infty} \left| x \frac{n}{n+1} \right| = \left| x \right| \lim_{n \to \infty} \frac{n}{n+1} = \left| x \right|. \]

Step 4: Determine the Convergence Set

According to the Absolute Ratio Test, the series converges if the limit found in Step 3 is less than 1: \( \left| x \right| < 1. \) This means the power series converges for the interval: \(-1 < x < 1.\)

Step 5: Include the Interval Points (Optional)

To determine if the series converges at \( x = -1 \) and \( x = 1 \), we need to test these points separately:- **For \( x = 1 \):** The series becomes \( 1 - 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \cdots \), which is an alternating harmonic series and converges.- **For \( x = -1 \):** The series becomes \( 1 - (-1) - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} - \cdots \), which does not converge.Thus, the convergence set is \( [-1, 1) \).

Key Concepts

Convergence Set

The convergence set of a power series is the set of values for which the series converges to a finite sum. For the given series, the convergence set is about finding for which values of \( x \) the series converges. The power series provided, \( 1 - x + \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4} - \cdots \), is characterized by alternating signs. This type of series, where terms switch between positive and negative, has special properties that help in determining convergence. To find the convergence set, we first identify the general term of the series, which is \((-1)^n\frac{x^n}{n}\). This identification is crucial as it allows us to apply convergence tests properly. For this series, using the Absolute Ratio Test (discussed below), we discover that the series converges for values \(-1 < x < 1\). This is our interval of convergence. To fully determine the convergence set, though, we must consider the behavior of the series at the endpoints \(x = -1\) and \(x = 1\). In this case: - At \(x = 1\), the series becomes an alternating harmonic series, which converges.- At \(x = -1\), the series diverges. Therefore, the complete convergence set for this power series is \([-1, 1)\).

Absolute Ratio Test

The Absolute Ratio Test is a tool in calculus used to determine the radius of convergence of a power series. By examining the behavior of the ratio of successive terms, it helps us understand whether the series converges or diverges when taken to an infinite length. For a power series like \(\sum a_n x^n\), the test involves evaluating the limit:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]In this particular problem, substituting the general term \(a_n = \frac{(-1)^n x^n}{n}\), the ratio becomes:\[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x}{1 + \frac{1}{n}} \right|\]As \(n\) approaches infinity, the expression simplifies to \(|x|\). For convergence, this limit must be less than 1, determining the interval \(-1 < x < 1\). This calculation reveals an important characteristic: the radius of convergence is 1. This radius tells us how far from the center of the series (typically at \(x=0\)) we can move while maintaining convergence.

Alternating Series

An alternating series is a sequence of terms in which the sign changes alternately between positive and negative as we move along the series. The presence of alternating signs affects convergence properties and allows us to apply the Alternating Series Test. The series \(1 - x + \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4} - \cdots\) is identified as an alternating series, with the general term \((-1)^n\frac{x^n}{n}\). This ultimately means that every other term subtracts in value followed by an addition. Alternating series have the potential to converge even when their terms do not individually approach zero if they satisfy specific criteria. By the Alternating Series Test, convergence is guaranteed if two conditions are met:

• The absolute value of the terms decreases monotonically.
• The limit of the terms, as the series progresses to infinity, approaches 0.

Our given series meets both these conditions as \(n\) increases, each term becomes smaller, and eventually falls to zero. Therefore, this aspect of alternating series helps in establishing convergence within the defined interval.